SST and SLP

Changheng Chen

Note: x = sst; y = slp.

2. Fields means and variances: space, time, spacetime

1. The GRAND (space+time) mean of my x and y: 27.5743 °C, 1.0110E3 hPa.
2. The GRAND (space+time) variance of my x and y: 2.9325 (°C)2, 2.3722 (hPa)2.
3. The GRAND standard deviations are: 1.7125 °C,21.5402 hPa.
4. The SPATIAL variances of my TIME MEAN longitude section are: 1.8427 (°C)2, 1.1617 (hPa)2.
5. The TEMPORAL variances of my LONGITUDE MEAN time series are: 0.3616 (°C)2, 0.5488 (hPa)2.

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3. (the assignment part)

1. Confirm that the time mean of the anomalies as defined above is 0. Yes.
2. Is the spatial mean of the anomalies (as defined above) 0? No.

__ _ _ Is it the same as the time series of the spatial mean of the raw data? Or is it a new thing?
_ _ _ _ _ _ Different things but related: (spatial mean of SST climate anomaly) = (spatial mean of SST) - (seasonal cycle of SST)

_ _ 3. My CLIMATOLOGICAL ANNUAL CYCLE have variance:

• 2.4853 (°C)2 (SST)
• 1.8683 (hPa)2 (SLP)

___4. My INTERANNUAL ANOMALY ARRAYS have variance:

• 0.4473 (°C)2 (SST)
• 0.5037 (hPa)2 (SLP)

_. _5. Fill out a variance decomposition table for field 1: feel free to add columns if you can define other parts.

Variance / (°C)2	SST
a) total variance of x	2.9325
b) purely spatial (variance of TIME mean at each lon)	1.8427
c) variance of (x minus its TIME mean at each lon)	1.0898
d) purely temporal (variance of LON mean at each time)	0.3616
e) variance of (x minus its LON mean at each time)	2.5710
f) remove both means (space-time variability)	0.7282
g) mean seasonal cycle	2.4853
h) deseasonalized anomalies	0.4473
i) variance of longitudinal mean of h	0.1025
j) h minus i	0.3448

_ _ 6. Discuss your results:

• (a) = (b) + (c)
• (a) = (d) + (e)
• (a) = (b) + (d) + (f)
• (a) = (c) + (e) - (f)
• (a) = (g) + (h)

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4. Further decomposition of anomx by scale (using rebinning).

What space and time scales (units: degrees and months) have the most variance in your anomx field? Create a contour plot of anomx -- can you see these "characteristic" scales by eye?

• SST field:
  • a scale factor greater than 30 (grid is ~75°) in longitude + a small scale factor in time --> largest gradient. No reasonable physical meaning is expected. Purely statistical result?
  • low scale factors in both longitude and time --> small gradient in variance. Variance depends more on large scale structures.

• SLP field:
  • low scale factors in both longitude and time --> small gradient in variance.
  • With scale factor (small) fixed in time, an increasing in scale factor in longitude does not affect the gradient very much.

Note: the above plots are in ln/ln scales.

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5. Scatter plot, correlation and covariance, regression-explained variance

1. Based on your data fields ), make subsets of your 2 variables x and y and make a scatter plot of these showing the strongest (positive or negative) correlation of one field with the other you can find. The subset might simply be all (x,t) values if your fields are very similar (olr, precip), or maybe the 240 time values at one longitude, or 144 longitudinal values in the time mean, or time series at different longitudes if some variability is offset in your two fields (like pressure and wind).

Time mean sst has a negative ralationship with time mean slp, especially from the longitude 170 to 257.5 degree.

A clear linear relationship between time mean sst and time mean slp shows up from longitude 170 to 257.5 degree.

__2. Now consider the covariance and correlation of the two subset arrays entering your scatterplot.

• What is the correlation coefficient corresponding to this scatter plot?
  -0.7102
• What are the standard deviations of your two data subsets?
  sst: 0.2862 °C; slp: 0.2570 hPa.
• What fraction of the variance of y can be 'explained' by linear regression on x (y = mx + b)? How does this relate to rho? How much y variance is explained? (variance: with units of y squared) What is m? Hint: these are simple questions: use the math formula, not a computer code (Hsieh section 1.4.2, Eq. 1.33).
  50.44%; fraction=rho^2; m=rho*std(y)/std(x)=-0.8945.
• What fraction of the variance of x can be 'explained' by linear regression on variable y? (x = ny + a)? How does this relate to rho? What is n? Hint: these are simple questions, use the math formula not computer code.
  n=1/m (i.e.: y = mx +b --> x = (1/m)*y - b/m ?).

__3. Now add uncorrelated (random) noise with variance 1 to one of your variables. This might be like observation error.

• How did the variance of y change when this noise was added?
  var(y(:))=1.1698 (hPa)2; var(noisey(:))=2.0899 (hPa)2.
• How did the correlation change?
  rho = corrcoef(x(:), noisey(:)) = -0.5089
• How do these changes affect the regression of y on x? How much (y+noise) variance is explained by linear regression on x? What is the new value of m in the new (noisey = mx + b) regressio?
  25.90%; m=-0.4795.
  • Hint: all these could be answered without using the computer, but it may help to confirm with data.

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6. Lagged correlation, covariance, and cross-covariance: questions

1. Show the zero-lag temporal covariance and correlation structures for your primary field. Interpret the results.

• The largest variance of sst shown in the first two plots goes from ~180° to ~300° in longitude, while variance at other locations is relatively small.
• The most interesting signal in the last plot is the negative correlation between 130°-160° (i.e. 130°E-160°E, west Pacific) and 180°-240° (i.e. 180°-60°W, middle and east Pacific). That is, sst anomaly varies in opposite way in east and west Pacific; it probably indicates El Nino/La Nina events.

%% 6 Lagged correlation, covariance, and cross-covariance
% deseasonalized anomalies
load sst_climate.mat; x = sst_climate';
load slp_climate.mat; y = slp_climate';
x = x-repmat(mean(x,2),1,240); % not necessary, since mean(x,2)=0 already.
y = y-repmat(mean(y,2),1,240);
 
MAXLAG = 24; % out to +/- 24 months
NLAGS = 2*MAXLAG + 1; % 49 lags are considered, including 0!!!
lags = linspace(-MAXLAG, MAXLAG, NLAGS);
 
NTSTATS = 240. - NLAGS; % trimmed off ends, so sum is over 191=N-1 months, not 240
for ilag = 1:NLAGS % lag index 1-49
    timelag = ilag-MAXLAG-1; % actual time lag, -24 to 24 months (-24: y is 24 months ahead of x)
    lagcovxy_temporal(:,:,ilag) = x(:,MAXLAG+1:end-MAXLAG) * y(:,MAXLAG+timelag+1:end-MAXLAG+timelag)'/NTSTATS;
    lagcovxx_temporal(:,:,ilag) = x(:,MAXLAG+1:end-MAXLAG) * x(:,MAXLAG+timelag+1:end-MAXLAG+timelag)'/NTSTATS;
    lagcovyy_temporal(:,:,ilag) = y(:,MAXLAG+1:end-MAXLAG) * y(:,MAXLAG+timelag+1:end-MAXLAG+timelag)'/NTSTATS;
end
 
stdx = repmat(std(x')',1,144);
stdy = repmat(std(y')',1,144);
 
% Make a few plots
figure(1)
contourf(lagcovxx_temporal(:,:,25)) % zero-lag covariance
title('zero-lag COVARIANCE of anomx = deseasonalized sst anomalies')
figure(2)
plot(stdx)
title('SST stdev as a function of lon')
figure(3)
contourf(lagcovxx_temporal(:,:,25)./(stdx.*stdx')) % zero-lag covariance, //NOT STRICTLY CORRECT MATHEMATICALLY//
title('zero-lag CORRELATION of anomx = deseasonalized sst anomalies')
 

__2. Show longitude-lag sections of the covariance or correlation of this field, for a base point at some longitude of interest.

• The above figure shows the covariance between sst anomaly (ssta) at longitude 267.5° (i.e., 87.5°W, east Pacific) and ssta at all longitudes as a function of lag. For a lag>10 months and lag<-10 months, the covariance between ssta at 87.5°W and ssta at 20°W-60°W is negative, however, it shows a clear positive covariance for a lag less than 6 months.

__3. Intepret the results in terms of the characteristic space and time scales of your anomalies. Can you see these characteristic scales in your original raw data?

• SST anomaly at eastern Pacific has a cycle of about less than two year. Other regions do not show such strong signal. This cycle can also be seen in the raw data.

__4. Share a longitude-lag slice of your lagged co-variance matrix for your TWO fields. Label it, interpret it.

• The above figure shows the covariance between time series of slp at 87.5°W (i.e. eastern Pacific) and those of sst at all longitudes as a function of lag. When lag<0, sst leads slp. For a lag of 2 years, an increase of slp at eastern Pacific two years ahead is most likely followed by an increase of sst in the same region. When lag>0, slp leads sst. The same conclusion can be drawn, however, the correlation shown in the above figure is relatively weak. On the other hand, the correlation is negative for a lag less than 6 months.